1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Projection speed mechanics

  1. Aug 13, 2009 #1
    i have to answer this question for an assignment that I need to do for mechanics. I am really really stuck - would somebody please mind helping....

    A ball thrown at an angle α to the horizontal just clears a wall. The horizontal and vertical distances to the top of the wall are X and Z respectively, with X.tan α>Z. Show that the projection speed is...

    X/cos α { g/(2(X.tanα – Z)}^1/2

    Where g is acc due to gravity. Determine the position of the highest point reached in terms of X. Z and tan α.
     
    Last edited by a moderator: May 11, 2014
  2. jcsd
  3. Aug 13, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Re: +++++++please help+++++++++++

    Hint: Write equations for the vertical and horizontal positions as functions of time.
     
  4. Aug 13, 2009 #3
    Re: +++++++please help+++++++++++

    Thanks - I have already done that but still nothing....
     
  5. Aug 13, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Re: +++++++please help+++++++++++

    Show what you did. Make use of the fact that when x = X, y = Z.
     
  6. Aug 13, 2009 #5
    Re: +++++++please help+++++++++++

    I have done that -
    i'm not sure which equation to use...

    I know that tx = ty so i have tried rearranging things and substituting. but i still have v's left in everything....

    i dont understand where the tan has come from...

    I'm guessing that i have to use v^2 = u^2 + 2as as the final eqn has a 2 in it but i dont understand how the two has got to the bottom of the fraction sign. I also dont understand where the squre root has come from.......
     
  7. Aug 13, 2009 #6
    Re: +++++++please help+++++++++++

    I dont even know if i'm supposed to be using calculus or not - this is supposed to be an as level mechanics problem
     
  8. Aug 13, 2009 #7

    Doc Al

    User Avatar

    Staff: Mentor

    Re: +++++++please help+++++++++++

    Again I ask you to write down the expressions for horizontal (x) and vertical (y) positions as a function of time. (They will also involve V and α.) The tan and the 2 will fall into place automatically when you later set up and solve those equations.

    (Nothing to do with calculus, just a little algebra.)
     
  9. Aug 13, 2009 #8
    Re: +++++++please help+++++++++++

    xcos(α) = (U + V)/2 * t , zsin(α) = (u + V)/2 * t
     
  10. Aug 13, 2009 #9
    Re: +++++++please help+++++++++++

    or x = [(ucos(α) + V/)2] *t and z = [(usin(α) + V)/2] * t
     
  11. Aug 13, 2009 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Re: +++++++please help+++++++++++

    Those aren't correct. Review the kinematic equations here: https://www.physicsforums.com/showpost.php?p=905663&postcount=2"

    (The ones you want relate displacement and time.)

    Getting closer, but still not right.
     
    Last edited by a moderator: Apr 24, 2017
  12. Aug 13, 2009 #11
    Re: +++++++please help+++++++++++

    I'm sorry but i still really dont understand what i am supposed to do with these.
     
  13. Aug 13, 2009 #12

    Doc Al

    User Avatar

    Staff: Mentor

    Re: +++++++please help+++++++++++

    Start with this general equation:
    [tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]

    How would you apply it to the horizontal and vertical motion of your projectile?
     
  14. Aug 13, 2009 #13
    Re: +++++++please help+++++++++++

    ok so s = ut + 1/2at^2.....

    x = ucos(α)t , Z = usin(α)t 1/2gt^2 but i'm still unsure about what i am supposed to do next...I already tried rearranging the X one in terms of t then substituting it into the other....which didn't work...if i divide them then i loose the cos(α) using pythagoras was still wrong...?
     
  15. Aug 13, 2009 #14
    Re: +++++++please help+++++++++++

    sorry thats supposed to be z = u.sin(α) - 1/2gt^2
     
  16. Aug 13, 2009 #15
    Re: +++++++please help+++++++++++

    i've made a substitution.....

    x/ucos9α) = t

    Z = u.sin(α).(x/cos(α)) - 1/2g[(x/ucos(α))^2]

    is this correct?
     
  17. Aug 13, 2009 #16

    Doc Al

    User Avatar

    Staff: Mentor

    Re: +++++++please help+++++++++++

    Try it again. (It worked for me. :wink:) Take the X equation and solve for t, then substitute that into the Z equation.

    Good. But you left out a t in the first term on the right hand side.
     
  18. Aug 13, 2009 #17

    Doc Al

    User Avatar

    Staff: Mentor

    Re: +++++++please help+++++++++++

    Almost. You dropped a u in the second term. But you are on the right track.
     
  19. Aug 13, 2009 #18
    Re: +++++++please help+++++++++++

    i get -

    Z = usinα.(x/cosα) - 1/2g[x/ucosα]^2 but now i have a cos^2 which i'm not sure how to get rid of??
    Thanks
     
  20. Aug 13, 2009 #19
    Re: +++++++please help+++++++++++

    re arranging i get -

    u^2 = xthanα - 1/2.g.(x^2/zcos^2α)

    Which is still incorrect, i really dont know where i am going wrong sorry
     
  21. Aug 13, 2009 #20

    Doc Al

    User Avatar

    Staff: Mentor

    Re: +++++++please help+++++++++++

    (1) Fix that second term--you left out a "u".
    (2) You don't want to "get rid of the cos^2". Just expand that square.
    (3) Simplify as much as possible, then solve for u. (Hint: Solve for 1/u^2 first.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Projection speed mechanics
  1. Mechanics (speed) (Replies: 3)

  2. Wind Speed Project (Replies: 8)

Loading...